Recent content by Gerardum

Its okay, I already handed it in. I messed up though. It turns out I just needed to find out that the two energy functions were degenerate and that means they are an eigenfunction of the hamiltonian.

So I ended up getting the function: -50*sin(Pi*x)*Pi^6*sin(Pi*y)*sin(5*Pi*z)-1458*sin(3*Pi*x)*Pi^6*sin(3*Pi*y)*sin(3*Pi*z) after applying the Hamiltonian to the Psi functions, but the two Psi terms now have different coefficients. Does that mean that Psi1 + Psi 2 is not an Eigenfunction?

But if sin(n*pi)=0, then wouldn't the whole psi function = 0? In that case it wouldn't be an eigenvalue, unless its a trivial solution. Do you think its only giving Psi to list the quantum numbers and I just need to solve for E and then list the result in terms of an eigenvalue?

The numbers correspond to the n values. (aka ψ1,1,5 = nx = 1, ny = 1, nz=5) What I am confused about is how to solve this equation if I am given the boundries in terms of L and not in terms of an actual unit of length, should I just assume that L=1?

1. Consider a particle of mass m in a cubic (3-dimensional) box with V(x,y,z) = 0 for 0 < x < L, 0 < y < L, and 0 < z < L and V(x,y,z) = ∞elsewhere. Is 1/\sqrt{2} * (ψ(1,1,5)+ψ(3,3,3)) an eigenfunction of the Hamiltonian for this system? If so, what is the eigenvalue? Explain your reasoning 2...